Proof of the first isomorphism theorem

Proof

1. Let $\varphi : G \to H$ be a group homomorphism with kernel $K$. $K$ is a normal subgroup of $G$, so the quotient group $G/K$ exists.
2. Define $\psi : G/K \to \varphi(G)$ by $\psi(gK) = \varphi(g)$.
3. The map $\psi$ is well defined. If $gK = hK$, then $g^{-1}h \in K$, so $\varphi(g)^{-1}\varphi(h) = e$ and $\varphi(g) = \varphi(h)$.
4. The map $\psi$ is a homomorphism, because $\psi(gK\,hK) = \psi(ghK) = \varphi(gh) = \varphi(g)\varphi(h)$.
5. The map $\psi$ is onto the image $\varphi(G)$, because each $\varphi(g)$ is $\psi(gK)$.
6. If $\psi(gK) = e$, then $\varphi(g) = e$, so $g \in K$ and $gK = K$. Thus the kernel of $\psi$ is trivial, and $\psi$ is injective.
7. Thus $\psi$ is an isomorphism, and $G/K \cong \varphi(G)$.

Idea

Send each coset of the kernel to the common image of its elements. This map is well defined, respects products, and has a trivial kernel.